Proof. (i) By Theorem 3.4.2(i), \(\mathbf {e}^\top \mathbf {e} = \mathbf {y}^\top (\mathbf {I} - \mathbf {H})^\top (\mathbf {I} - \mathbf {H})\mathbf {y} = \mathbf {y}^\top (\mathbf {I} - \mathbf {H})\mathbf {y}\). And \(\mathbf {y}^\top \mathbf {H}\mathbf {y} = \mathbf {y}^\top \mathbf {X}\mathbf {C}\mathbf {X}^\top \mathbf {y} = \mathbf {y}^\top \mathbf {X}\hat {\boldsymbol \beta } = \hat {\boldsymbol \beta }^\top \mathbf {X}^\top \mathbf {y}\).

(ii) By Lemma 3.1.4 with \(\mathbf {A} = \mathbf {I} - \mathbf {H}\), \(\boldsymbol \mu = \mathbf {X}\boldsymbol \beta \) and \(\mathbf {V} = \sigma ^2\mathbf {I}\), \[ E(SS_{\mathrm {Res}}) = \sigma ^2\operatorname {tr}(\mathbf {I} - \mathbf {H}) + \boldsymbol \beta ^\top \mathbf {X}^\top (\mathbf {I} - \mathbf {H})\mathbf {X}\boldsymbol \beta = \sigma ^2(n - p) + 0 , \] by Theorem 3.4.2(iv) and (ii). □

Solution. From the residuals, \(SS_{\mathrm {Res}} = 1 + 1 + 1 + 1 + 0 + 0 = 4\). Check with Theorem 3.6.1(i): \(\mathbf {y}^\top \mathbf {y} = \sum y_i^2 = 887\) and \[ \hat {\boldsymbol \beta }^\top \mathbf {X}^\top \mathbf {y} = 3(69) + 2(244) + 1(188) = 207 + 488 + 188 = 883, \] so \(SS_{\mathrm {Res}} = 887 - 883 = 4\). With \(n - p = 6 - 3 = 3\), \[ \hat \sigma ^2 = \frac 43 = 1.3333 . \] The diagonal of \(\mathbf {C}\) is \(C_{00} = 510/378\), \(C_{11} = 33/378\), \(C_{22} = 96/378\), so \begin {align*} \operatorname {se}(\hat \beta _0) &= \sqrt {1.3333 \times 510/378} = 1.341, \\ \operatorname {se}(\hat \beta _1) &= \sqrt {1.3333 \times 33/378} = 0.341, \\ \operatorname {se}(\hat \beta _2) &= \sqrt {1.3333 \times 96/378} = 0.582 . \end {align*}

The estimated covariance of \(\hat \beta _1\) and \(\hat \beta _2\) is \(\hat \sigma ^2 C_{12} = 1.3333 \times (-30/378) = -0.106\). It is negative because, in these data, fertiliser and rainfall tend to rise together: credit for a high yield can be shifted from one to the other, so an overestimate of one coefficient tends to go with an underestimate of the other.